229. Torque on a Current Loop: Motors and MetersLearning Objectives
Let us examine the force on each segment of the loop in Figure 1 to find the torques produced about the axis of the vertical shaft. (This will lead to a useful equation for the torque on the loop.) We take the magnetic field to be uniform over the rectangular loop, which has width $w$ and height $l$. First, we note that the forces on the top and bottom segments are vertical and, therefore, parallel to the shaft, producing no torque. Those vertical forces are equal in magnitude and opposite in direction, so that they also produce no net force on the loop. Figure 2 shows views of the loop from above. Torque is defined as $\tau =\text{rF}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta $, where $F$ is the force, $r$ is the distance from the pivot that the force is applied, and $\theta $ is the angle between $r$ and $F$. As seen in Figure 2(a), right hand rule 1 gives the forces on the sides to be equal in magnitude and opposite in direction, so that the net force is again zero. However, each force produces a clockwise torque. Since $r=w/2$, the torque on each vertical segment is $(w/2)F\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta $, and the two add to give a total torque. $$\tau =\frac{w}{2}F\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta +\frac{w}{2}F\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =\text{wF}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta $$Now, each vertical segment has a length $l$ that is perpendicular to $B$, so that the force on each is $F=\text{IlB}$. Entering $F$ into the expression for torque yields $$\tau =\text{wIlB}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta .$$If we have a multiple loop of $N$ turns, we get $N$ times the torque of one loop. Finally, note that the area of the loop is $A=\text{wl}$; the expression for the torque becomes $$\tau =\text{NIAB}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta .$$This is the torque on a currentcarrying loop in a uniform magnetic field. This equation can be shown to be valid for a loop of any shape. The loop carries a current $I$, has $N$ turns, each of area $A$, and the perpendicular to the loop makes an angle $\theta $ with the field $B$. The net force on the loop is zero. Example 1: Calculating Torque on a CurrentCarrying Loop in a Strong Magnetic FieldFind the maximum torque on a 100turn square loop of a wire of 10.0 cm on a side that carries 15.0 A of current in a 2.00T field. Strategy Torque on the loop can be found using $\tau =\text{NIAB}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta $. Maximum torque occurs when $\theta =\text{90\xba}$ and $\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =1$. Solution For $\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =1$, the maximum torque is $${\tau}_{\text{max}}=\text{NIAB}.$$Entering known values yields $$\begin{array}{lll}{\tau}_{\text{max}}& =& \left(\text{100}\right)\left(\text{15.0 A}\right)\left(\text{0.100}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{2}\right)\left(2\text{.}\text{00 T}\right)\\ & =& \text{30.0 N}\cdot \mathrm{m.}\end{array}$$Discussion This torque is large enough to be useful in a motor. The torque found in the preceding example is the maximum. As the coil rotates, the torque decreases to zero at $\theta =0$. The torque then reverses its direction once the coil rotates past $\theta =0$. (See Figure 2(d).) This means that, unless we do something, the coil will oscillate back and forth about equilibrium at $\theta =0$. To get the coil to continue rotating in the same direction, we can reverse the current as it passes through $\theta =0$ with automatic switches called brushes. (See Figure 3.)
Section Summary
Conceptual QuestionsExercise 1Draw a diagram and use RHR1 to show that the forces on the top and bottom segments of the motor’s current loop in Figure 1 are vertical and produce no torque about the axis of rotation. Problems & ExercisesExercise 1(a) By how many percent is the torque of a motor decreased if its permanent magnets lose 5.0% of their strength? (b) How many percent would the current need to be increased to return the torque to original values? Show/Hide Solution Solution(a) $\text{\tau}$ decreases by 5.00% if B decreases by 5.00% (b) 5.26% increase Exercise 2(a) What is the maximum torque on a 150turn square loop of wire 18.0 cm on a side that carries a 50.0A current in a 1.60T field? (b) What is the torque when $\theta $ is $\text{10}\text{.}\mathrm{9\xba?}$ Exercise 3Find the current through a loop needed to create a maximum torque of $9\text{.}\text{00 N}\cdot \text{m.}$ The loop has 50 square turns that are 15.0 cm on a side and is in a uniform 0.800T magnetic field. Show/Hide Solution Solution10.0 A Exercise 4Calculate the magnetic field strength needed on a 200turn square loop 20.0 cm on a side to create a maximum torque of $\text{300 N}\cdot \text{m}$ if the loop is carrying 25.0 A. Exercise 5Since the equation for torque on a currentcarrying loop is $\tau =\text{NIAB}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta $, the units of $\mathrm{N}\cdot \mathrm{m}$ must equal units of $\mathrm{A}\cdot {\mathrm{m}}^{2}\phantom{\rule{0.25em}{0ex}}\mathrm{T}$. Verify this. Show/Hide Solution Solution$A\cdot {m}^{2}\cdot T=A\cdot {m}^{2}\left(\frac{N}{A\cdot m}\right)=N\cdot m$. Exercise 6(a) At what angle $\theta $ is the torque on a current loop 90.0% of maximum? (b) 50.0% of maximum? (c) 10.0% of maximum? Exercise 7A proton has a magnetic field due to its spin on its axis. The field is similar to that created by a circular current loop $0\text{.}\text{650}\times {\text{10}}^{\text{15}}\phantom{\rule{0.25em}{0ex}}\mathrm{m}$ in radius with a current of $1\text{.}\text{05}\times {\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}\mathrm{A}$ (no kidding). Find the maximum torque on a proton in a 2.50T field. (This is a significant torque on a small particle.) Show/Hide Solution Solution$3\text{.}\text{48}\times {\text{10}}^{\text{26}}\phantom{\rule{0.25em}{0ex}}\mathrm{N}\cdot \mathrm{m}$ Exercise 8(a) A 200turn circular loop of radius 50.0 cm is vertical, with its axis on an eastwest line. A current of 100 A circulates clockwise in the loop when viewed from the east. The Earth’s field here is due north, parallel to the ground, with a strength of $3\text{.}\text{00}\times {\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\mathrm{T}$. What are the direction and magnitude of the torque on the loop? (b) Does this device have any practical applications as a motor? Exercise 9Repeat Exercise 1, but with the loop lying flat on the ground with its current circulating counterclockwise (when viewed from above) in a location where the Earth’s field is north, but at an angle $\text{45}\text{.}\mathrm{0\xba}$ below the horizontal and with a strength of $\text{6.}\text{00}\times {\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\mathrm{T}$. Show/Hide Solution Solution(a) $\text{0.666 N}\cdot \mathrm{m}$ west (b) This is not a very significant torque, so practical use would be limited. Also, the current would need to be alternated to make the loop rotate (otherwise it would oscillate).
